Understanding y'=2/3x-1/3(x2-4)+x2/3*2x

  • Thread starter yoleven
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In summary, the conversation discusses the simplification of a derivative equation with the input y=x2/3(x2-4). One person provides two different expressions for the derivative, while the other questions how the numerator of the simplified equation was obtained. Clarification is needed on the specific variables and terms used in the equations.
  • #1
yoleven
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1

Homework Statement


y=x2/3(x2-4)



Homework Equations





The Attempt at a Solution


1.y'=2/3x-1/3(x2-4)+x2/3*2x

2.y'=2x2-8+2x5/2/3(x1/3)

the book agrees with my unsimplified (#1)derivative but it goes further to simplify the equation as
y'= 8x2-8/3(x1/3)

I can't see how they got 2(x2-4)+x2/3*2x to be 8x2-8
in the numerator.
Would someone explain this to me?
 
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  • #2
It's difficult to tell in your second expression what's in the numerator and what's in the denominator.

Is 8x^2 - 8 in the numerator?
Are both 3 and x^(1/3) in the denominator?

I can't explain it to you if I don't know what you wrote.
 

What is the equation "y'=2/3x-1/3(x2-4)+x2/3*2x" used for?

The equation "y'=2/3x-1/3(x2-4)+x2/3*2x" is used to represent the derivative of a function. It shows the rate of change of the function with respect to the variable x.

What do the different parts of the equation "y'=2/3x-1/3(x2-4)+x2/3*2x" represent?

The "y'" represents the derivative of the function, while "2/3x" and "x2/3*2x" represent the different terms in the function. The term "1/3(x2-4)" represents the derivative of the inside function of the parentheses.

How do you solve an equation like "y'=2/3x-1/3(x2-4)+x2/3*2x"?

To solve an equation like this, you can use the power rule and sum/difference rule for derivatives. You would first simplify the equation by distributing the constants and using the power rule to find the derivative of each term. Then, you would combine like terms and simplify the equation further.

What is the significance of the constant terms in "y'=2/3x-1/3(x2-4)+x2/3*2x"?

The constant terms in this equation represent the slope of the function at any given point. The constant term of "2/3" represents the slope of the linear term "2/3x", while the constant term of "1/3" represents the slope of the quadratic term "1/3(x2-4)".

Can the equation "y'=2/3x-1/3(x2-4)+x2/3*2x" be used to find the tangent line of a function?

Yes, the equation "y'=2/3x-1/3(x2-4)+x2/3*2x" can be used to find the tangent line of a function. The slope of the tangent line is equal to the value of the derivative at a specific point on the function.

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